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The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?
 one year ago
 one year ago
The x coordinate of a point on the curve 9y^2 = x^3 , the normal at which cuts off equal intercept on the axes is ?
 one year ago
 one year ago

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sauravshakyaBest ResponseYou've already chosen the best response.1
9y^2=x^3 d(9y^2)/dx =d(x^3)/dx 18y dy/dx =3x^2 dy/dx =x^2/(6y)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
The gradient of tangent is x^2/(6y) So, gradient of normal is 6y/(x^2)
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
NOW, let xintercept of the normal be a then yintercept is also a.
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Gradient of normal = (a0)/(0a) 6y/(x^2) =1
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Now,dw:1359201825424:dw and solve for x
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
dw:1359201971611:dw
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
U need to solve that
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
6y/(x^2) =1......then wat did u do ?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.1
Substitute y interms of x
 one year ago
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